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I have recently started to learn algebraic geometry and this question has been bugging me.

An affine variety is a zero set of a collection of polynomials in affine space and a projective variety is a zero set of a collection of polynomials in projective space.

Affine space has the property that a group action is available on it while projective space has the property that it is compact.

So one of the two definitions comes with something algebraic in addition while the other comes with something topological.

But it is not clear to me what the advantages of either are over the other. Compactness always makes things easier so it would not surprise me if it would make things easier in algebraic geometry, too. But exactly how it does is not clear to me.

I guess it will depend on which branch of algebraic geometry, too.

So let's assume, just for the purpose of this question, that I want to study singularities of varieties.

Please could someone explain to me, perhaps even provide a simple example, what the advantages of one type of variety is over the other? How to use the group action or the compactness to classify singularities?

I think once I see examples it will all become very clear.

student
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  • Have you seen the definition of a variety in general? – Tobias Kildetoft May 20 '16 at 10:48
  • @TobiasKildetoft No, the book by Karen Smith so far is about affine varieties only. After glancing at Wikipedia it appears that there are many different ways to define abstract varieties. – student May 20 '16 at 12:30
  • (1) $\mathbb P^n$ has an action of $PGL_n$. What group action on $\mathbb A^n$ are you talking about? (2) The usefulness of compactness (globally) is subtle already in differential geometry, so it should make sense that it takes some time to understand in this setting as well. – Hoot May 20 '16 at 15:27
  • @Hoot The definition of affine space states that an affine space comes endowed with a free and transitive action. – student May 21 '16 at 01:46
  • Well, I've given you a free and transitive action on $\mathbb P^n$, so this cannot be counted as a win for $\mathbb A^n$, I don't think. – Hoot May 21 '16 at 02:01
  • See this question, which has a nice answer: Bezout's Theorem. – Tomi Tyrrell May 21 '16 at 05:01

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